1. **Problem statement:** Find the shortest distance between the two parallel lines $l_1: y = 2x + 4$ and $l_2: 6x - 3y - 9 = 0$. 2. **Rewrite lines in standard form:** - Line $l_1$ is $y = 2x + 4$. - Rewrite as $2x - y + 4 = 0$. 3. **Check parallelism:** - Both lines have the form $Ax + By + C = 0$. - $l_1: 2x - y + 4 = 0$ - $l_2: 6x - 3y - 9 = 0$ - Note that $l_2$ coefficients are $3$ times those of $l_1$ except for the constant term, confirming they are parallel. 4. **Formula for shortest distance between two parallel lines:** $$ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} $$ where lines are $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$. 5. **Identify coefficients:** - $A = 2$, $B = -1$ - $C_1 = 4$ (from $2x - y + 4 = 0$) - $C_2 = -9/3 = -3$ (divide $l_2$ by 3 to match $l_1$ coefficients: $2x - y - 3 = 0$) 6. **Calculate distance:** $$ d = \frac{|(-3) - 4|}{\sqrt{2^2 + (-1)^2}} = \frac{7}{\sqrt{4 + 1}} = \frac{7}{\sqrt{5}} = \frac{7\sqrt{5}}{5} $$ 7. **Final answer:** The shortest distance between the lines $l_1$ and $l_2$ is $\boxed{\frac{7\sqrt{5}}{5}}$.